181 1.8 Absolute Value Equations and Inequalities Finally, x 73 is satisfied by all real numbers whose undirected distances from 0 are greater than 3. These numbers are less than -3 or greater than 3, so the solution set is 1-∞, -32 ´13, ∞2. Notice in Figure 19 that the union of the solution sets of x = 3, x 63, and x 73 is the set of real numbers. These observations support the cases for solving absolute value equations and inequalities summarized in the table that follows. If the equation or inequality fits the form of Case 1, 2, or 3, change it to its equivalent form and solve. The solution set and its graph will look similar to those shown. Solving Absolute Value Equations and Inequalities Absolute Value Equation or Inequality* Equivalent Form Graph of the Solution Set Solution Set Case 1: x = k x = k or x = -k –k k 5-k, k6 Case 2: x 6k -k 6x 6k –k k 1k, k2 Case 3: x 7k x 6 -k or x 7k –k k 1∞, -k2 ´1k, ∞2 *For each equation or inequality in Cases 1–3, assume that k 70. In Cases 2 and 3, the strict inequality may be replaced by its nonstrict form. Additionally, if an absolute value equation takes the form a = b , then a and b must be equal in value or opposite in value. Thus, the equivalent form of ∣ a∣ = ∣ b∣ is a =b or a =−b. Absolute Value Equations Because absolute value represents undirected distance from 0 on a number line, solving an absolute value equation requires solving two possibilities, as shown in the examples that follow. EXAMPLE 1 Solving Absolute Value Equations (Case 1 and the Special Case ∣ a∣ = ∣ b∣) Solve each equation. (a) 5 - 3x = 12 (b) 4x - 3 = x + 6 SOLUTION (a) For the given expression 5 - 3x to have absolute value 12, it must represent either 12 or -12. This equation fits the form of Case 1. 5 - 3x = 12 5 - 3x = 12 or 5 - 3x = -12 Case 1 -3x = 7 or -3x = -17 Subtract 5. x = - 7 3 or x = 17 3 Divide by -3. Check the solutions -7 3 and 17 3 by substituting them in the original absolute value equation. The solution set is E -7 3, 17 3 F. Don’t forget this second possibility.
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